Functions

Algebra Level 3

If the function $f : \mathbb{R} \rightarrow \mathbb{R} , f(x) = \cos(\dfrac{\pi}{2}[x])$ is an onto function, then the co-domain of this function is:-

\{-1, 0\}

[-1, 1)

[0, 1)

\{ -1, 0, 1\}

2 solutions

Ashish Menon
Apr 28, 2016

If [x] = ....,-6,-2,2,6,10,... then f (x) = -1.
If [x] = .....-3,,-1,1,3,5,...... then f (x) = 0.
If [x] = ...,-8,-4,0,4,8,12,.... then f (x) = 1.

So, for all real values of [x], the answer is either 1 or 0 or -1.
Since the function is one-to-one, the range is equal to the co-domain.

So, co-domain = $\boxed{\{-1, 0, 1\}}$

Do you mean the range of the function?

Mark C - 5 years, 1 month ago

Yes, the question was intended to troll the problem solvers. If the function is one-to-one, the co-domain is equal to its range. ;)

Ashish Menon - 5 years, 1 month ago

Gotcha. ;-)

Mark C - 5 years, 1 month ago

@Mark C – Great :+1:

Ashish Menon - 5 years, 1 month ago

@Ashish Menon – No. If the function is one to one, co-domain is not necessarily equal to range.
Example :
$f:\left[\dfrac{-\pi}{2},\dfrac{\pi}{2}\right]\to R , f(x) = \sin(x)$
is a one-to- function.

What you mean is "for the function to be surjective or onto what should be the co-domain?"The co-domain is equal to the range when the function is surjective and not injective.

A Former Brilliant Member - 5 years, 1 month ago

@A Former Brilliant Member – Oh wait, yes I really did mean onto function, but i wrote one - to - one XD Thanks I have changed it.

Ashish Menon - 5 years, 1 month ago

^In case of onto functions, the range and the co-domain are the same

Rifath Rahman - 1 year, 7 months ago

Jitu Mewada
May 3, 2017

y=cos(π\2[x]) Re Write it as 2/πcos^-y=[x] As cos^- is [0 π] so Therefore 2/πcos^-y ranges from [0 2] Values of x (0 2) There fore [x] ={1 0 2} As putting value of x we get 2/πcos^-y=0 y=0 2/πcos^-y=1 y=1 2/πcos^-y=2 y=-1 There fore y={-1 0 1} Any doubt msg me

Functions

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